A jar contains 2 red, 5 green, 2 blue and 6 yellow marbles. A marble is chosen at random from the jar. After replacing it, a second marble is chosen. What is the probability of choosing a green and then a yellow marble?​

Step-by-step explanation:

The probability of picking an item A in a box of N items depends on nA, the amount of items A in the box at the time of picking, with the kinda instinctive value:

p = nA / N

then it means that if you have no green ball, the probability is 0/N = 0, if you only have green balls, it means you have a probability of N/N = 1 = 100%, and if you have as many green balls as non-green balls, you have a probability of nA/(2nA) = 0.5 = 50%.

Here the probability of first picking a green ball (nG=5) in a box of N=3+5+2+6=16 for the 1st pick is:

pG1 = 5/16

Now the trick here is that once you picked the green ball, you put it back in the box before picking another ball. So you still have N=16 balls for the second pick, and the probability of picking a yellow ball (nY=6) for the 2nd pick is:

pY2 = 6/16 = 3/8

(it would have been slightly higher, at 6/15, if you did NOT put back the green marble in the box).

Now you want BOTH events to be true. So you want G1 to be true AND Y2 to be true. The two events are independent (the probability of Y2 is 6/16 if you had G1 but is also 6/16 if you failed G1), so the probability of having both is simply done by multiplicating them:

pG1Y2 = pG1 * pG2 = 15/128

468

step-by-step explanation:

the answer is -1 1/5, also a

hope i

-lavira : )